Verify property of inner product

I struggle with verifying positive-definiteness, in particular $$\langle f,f\rangle =0\implies f=0.$$ I know that for continuous non-negative functions, if the integral vanishes, then the function is identically ##0##. Here, however, ##f## being in ##L^2## does not make it continuous, right? This is from Bridge's Foundations of Real and Abstract Analysis.

That definition is confusing. Why should we care about continuity if all that counts is Lebesgue integrability?

I have a version in my book where everything is about continuity (not ##L_2##), and a version about Lebesgue integrability (##L_2##) without weight function ##w.## Your version looks like comparing apples and oranges to me.

E.g., we have by the mean value theorem that
$$
0=\bigl\langle f\, , \,f \bigr\rangle =\int_a^b w(t) \|f\|^2 \,dt = w(\xi)\int_a^b \|f\|^2 \,dt
$$
And here is the problem: if ##w(\xi)\neq 0## then we are done. But we cannot know that this is the case. If we remain in the box of oranges, and demand
$$
\int_a^b w(t)h(t)\,dt =0 \Longrightarrow h\equiv 0 \text{ for all } h\in L_2(I)
$$
instead of restricting us to continuous functions only, then it makes sense and we can conclude by setting ##h=\|f\|^2.## The other possibility is to require ##w## to be strictly positive. That would fill the gaps of potential Lebesgue Null sets, too.

psie said:

Homework Statement: Let ##w(t)## be a nonnegative continuous function on ##I=[a,b]## such that if ##f\in C(I)## and ##\int_I w(t)f(t) dt=0##, then ##f=0##. Prove that $$\langle f,g\rangle=\int_a^b w(t)f(t) \overline{g(t)} dt,$$ defines an inner product on ##L^2(I)##.
Relevant Equations: The properties of an inner product; conjugate symmetry, linearity in the first argument and positive-definiteness.

I struggle with verifying positive-definiteness, in particular $$\langle f,f\rangle =0\implies f=0.$$ I know that for continuous non-negative functions, if the integral vanishes, then the function is identically ##0##. Here, however, ##f## being in ##L^2## does not make it continuous, right? This is from Bridge's Foundations of Real and Abstract Analysis.

According to Wolfram, ##L^2## is actually not a set of functions, but of equivalence classes of functions, where functions that are identical except on sets of measure zero are equivalent.
Under that interpretation, we can conclude from ##\langle f, f\rangle = 0## that ##f## is zero except on a set of measure zero. That means that ##f## is in the same equivalence class as the zero function and hence is zero in the space ##L^2##.

andrewkirk said:

According to Wolfram, ##L^2## is actually not a set of functions, but of equivalence classes of functions, where functions that are identical except on sets of measure zero are equivalent.
Under that interpretation, we can conclude from ##\langle f, f\rangle = 0## that ##f## is zero except on a set of measure zero. That means that ##f## is in the same equivalence class as the zero function and hence is zero in the space ##L^2##.

That's the case if ##w\equiv 1.## But how do we exclude that ##L_2\ni w\equiv 0.##?

fresh_42 said:

That's the case if ##w\equiv 1.## But how do we exclude that ##L_2\ni w\equiv 0.##?

It is not ##w## but ##f## and ##g## that are indicated to be elements of ##L^2## in the OP.
##w## is the measure function used to define the inner product on the space. In the wolfram article I linked they call it ##\mu##.
If ##w\equiv 0## then all sets have ##w##-measure zero, so there is only one equivalence class of functions, and the inner product space becomes trivial.

andrewkirk said:

It is not ##w## but ##f## and ##g## that are indicated to be elements of ##L^2## in the OP.
##w## is the measure function used to define the inner product on the space. In the wolfram article I linked they call it ##\mu##.
If ##w\equiv 0## then all sets have ##w##-measure zero, so there is only one equivalence class of functions, and the inner product space becomes trivial.

It remains to show that ##\bigl\langle f\, , \,f \bigr\rangle = 0 ## implies ##f=0.## How is it done?

We must somehow rule out that ##w(t)=0## whenever ##f(t)\neq 0.## I see this if we require
a) ##w(t)>0## for all ##t\in I##
or
b) ##\int_I w(t)f(t)dt=0 \Rightarrow f=0## for all ##f\in L_2##

I do not see it if ##f\in C(I)## in condition b), a significantly smaller set of functions.

Just a remark.

The continuity requirement for the weight function ##w## makes sense because it defines a property for all ##f\in L_2## and not those depending on ##w.## The weight shouldn't disturb, and continuity guarantees that.

psie said:

Homework Statement: Let ##w(t)## be a nonnegative continuous function on ##I=[a,b]## such that if ##f\in C(I)## and ##\int_I w(t)f(t) dt=0##, then ##f=0##.

Isn't that impossible? Whatever ##w## you choose, can't you find some non-zero ##f## that makes the integral vanish?

psie said:

Homework Statement: Let ##w(t)## be a nonnegative continuous function on ##I=[a,b]## such that if ##f\in C(I)## and ##\int_I w(t)f(t) dt=0##, then ##f=0##.

Here's an outline proof that any such ##w \equiv 0## on ##[a, b]##

As ##w## is continuous and nonnegative on ##[a, b]##, then either ##w \equiv 0## or ##\int_a^b w(t) dt > 0##. Assume the latter.

We can find ##a < c < b##, such that:
$$\int_a^c w(t)dt = \int_c^b w(t) dt > 0$$Now, choose any continuous, non-negative function ##f_1## defined on ##[a, c]## such that ##f_1(c) = 0##. And any continuous non-positive function ##f_2## defined on ##[c, b]## such that ##f_2(c) = 0##. Also, neither ##f_1## nor ##f_2## is identically zero.

Now, we have:
$$\int_a^c w(t)f_1(t)dt > 0, \ \text{and} \ \int_c^b w(t)f_2(t) dt < 0$$Note that we can multiply ##f_2## by some constant so that:
$$\int_a^c w(t)f_1(t)dt = -\int_c^b w(t)f_2(t) dt$$Finally, the function ##f## defined as ##f_1## on ##[a, c]## and ##f_2## on ##[c, b]## is continuous, not identially zero and satisfies:
$$\int_a^b w(t)f(t)dt = 0$$Hence, there is no ##w## with the originally stated property.

fresh_42 said:

The other possibility is to require ##w## to be strictly positive. That would fill the gaps of potential Lebesgue Null sets, too.

I know the assumption are quite confusing, probably not right either, but why would a positive weight function make the statement ##\langle f,f\rangle\implies f=0## a.e. true?

psie said:

I know the assumption are quite confusing, probably not right either, but why would a positive weight function make the statement ##\langle f,f\rangle\implies f=0## a.e. true?

Do you mean ##\langle f, f\rangle = 0##? That implies that ##w(t)|f(t)|^2=0##, which means ##f=0## a.e.

Is this from a textbook?

martinbn said:

Is this from a textbook?

psie said:

This is from Bridge's Foundations of Real and Abstract Analysis.

I assume that it is a cut-and-paste error or a simple little mistake.

PeroK said:

Isn't that impossible? Whatever ##w## you choose, can't you find some non-zero ##f## that makes the integral vanish?

This is strange. I looked in the book and it is written exactly like that.

martinbn said:

This is strange. I looked in the book and it is written exactly like that.

Although I laboured through a proof, it's obvious there can be no such ##w##.

PeroK said:

Although I laboured through a proof, it's obvious there can be no such ##w##.

Matches with my textbook: ##w\ge 0## in the case of ##C(I)## and ##w\equiv 1## for ##L_2(I)##.

fresh_42 said:

Matches with my textbook: ##w\ge 0## in the case of ##C(I)## and ##w\equiv 1## for ##L_2(I)##.

The initial property is close to saying that the only continuous function whose integral is zero on a closed interval is the zero function. That's what it amounts to.

PeroK said:

The initial property is close to saying that the only continuous function whose integral is zero on a closed interval is the zero function. That's what it amounts to.

Well, I haven't found that strange condition, only the definitions of inner products on the two spaces.

PeroK said:

Although I laboured through a proof, it's obvious there can be no such ##w##.

No, i agree with you. It is strange that it is in the book.

May be ##C(I)## means continuous and positive. And positive means non-negative. This way ##w## is continuous non-negative and is zero on set of measure zero.

psie said:

Homework Statement: Let ##w(t)## be a nonnegative continuous function on ##I=[a,b]## such that if ##f\in C(I)## and ##\int_I w(t)f(t) dt=0##, then ##f=0##.

Consider functions in the form ##f_c(t)=c-t##

For ##c=a##, function ##f_{a}(t)=a-t \leq 0## , for all ##t \in [a,b]##.
For ##c=b+1##, function ##f_{b+1}(t)=b+1-t \geq 1## , for all ##t \in [a,b]##.

Function
$$F(c)=\int_a^b w(t) (c-t) dt$$
is continuous. For ##F(a) \leq 0## and ##F(b+1) \geq 0##.

There exists ##c_0 \in [a, b+1]## such that
$$F(c_0)=\int_a^b w(t) (c_0-t) dt=0$$
but ##f(t)=c_0-t## is not 0 for all ##t \in [a,b]##

I assume that ##f(t)## should also be a nonnegative continuous function.

PeroK said:

Here's an outline proof that any such ##w \equiv 0## on ##[a, b]##
...
$$\int_a^b w(t)f(t)dt = 0$$Hence, there is no ##w## with the originally stated property.

Now I see that I did the same thing in a similar way.

hmmm,... it seems as if there is no such function w, even if a=b.